## virtus Group Title find the equation of the locus of a point that moves so that its distance from the line 3x-4y+9=0 is always 5 2 years ago 2 years ago

1. ganeshie8 Group Title

let that point be P (x, y)

2. ganeshie8 Group Title

distance from a point to a line = $$\frac{Ax+By+C}{\sqrt{A^2 + B^2}}$$

3. ganeshie8 Group Title

can you put the equation for distance ? from point P(x, y) to line 3x-4y+9=0

4. virtus Group Title

oh so it would be (3x-4y+9)/square root( 3^2 +(-4)^2)

5. virtus Group Title

(3x -4y+9)/5 =5 3x -4y+9 =25 3x-4y -16 =0

6. virtus Group Title

is that right ganeshi8?

7. virtus Group Title

btw i thought there was an ABSOLUTE SIGN around Ax + By + C for the formula

8. ganeshie8 Group Title

yep! u r right.. we must get two lines

9. ganeshie8 Group Title

|(3x -4y+9)|/5 =5

10. ganeshie8 Group Title

3x-4y+9 = 25, and, 3x-4y + 9 = -25

11. virtus Group Title

ngawww i get it !

12. ganeshie8 Group Title

it tricked me.... i was thinking there should be two lines at the very end. nice catch :)

13. virtus Group Title

thanks again - LOL and about writing GAPS yesterday, it wasn't meant to be rude or anything, just expressing my gratitude for your help ;)

14. ganeshie8 Group Title

lol fine i get it nw :))

15. ganeshie8 Group Title

|dw:1341791339629:dw|

16. ganeshie8 Group Title

two lines we just got... which are at a distance of 5 from the given line... they look as above